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Let $X$ ba a smooth hyperelliptic curve of genus $g$, and let $f:X\rightarrow X$ be the hyperelliptic involution. Consider a $K3$ surface $S$ with an involution $g$ without fixed points. The quotient $F = S/g$ is an Eniques surface.

Now $f\times g:X\times S\rightarrow X\times S$ does not have fixed points, and the quotient $(X\times S)/(f\times g)\rightarrow F$ is a family of hyperelliptic curves parametrized by $E$ whose fibers are all isomorphic to $X$.

My question is: why isn't the family $(X\times S)/(f\times g)\rightarrow F$ trivial?

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2 Answers 2

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If it was, $Y:=(X\times S)/(f\times g)$ would be isomorphic to $X\times F$. One way to see this is not the case is to look at 3-forms: $X\times F$ has no nonzero holomorphic 3-forms (because $H^0(F,\Omega ^2_F)=0$). On the other hand, let $\omega $ be the generator of $H^0(S,\Omega ^2_S)$; for any $\alpha \in H^0(X,\Omega ^1_X)$, the form $pr_1^*\alpha \wedge pr_2^*\omega $ on $X\times S$ is invariant under $f\times g$, hence comes from a 3-form on $Y$.

This is a typical example of an isotrivial family : it is not trivial, but becomes trivial after the étale base change $S\rightarrow F$.

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  • $\begingroup$ Thanks a lot for the answer. I'm just missing one thing. Why is the 3-form $pr_1^*\alpha\wedge pr_2^{*}\omega$ on $X\times S$ invariant under $f\times g$ ? $\endgroup$
    – user58018
    Commented Sep 26, 2014 at 21:24
  • $\begingroup$ On the hyperelliptic curve $X$, there is no $f$-invariant 1-form (such a form would descend to $\mathbb{P}^1$), hence $f^*\alpha =-\alpha $ for all $\alpha \in H^0(X,\Omega ^1_X)$. For an analogous reason, $g^*\omega =-\omega $ on $S$. Thus $pr_1^*\alpha $ and $pr_2^*\omega $ are anti-invariant under $f\times g$, hence their product is invariant. $\endgroup$
    – abx
    Commented Sep 28, 2014 at 18:19
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Apart from the nice answer by abx, one can also argue as follows: the obvious commutative diagram
$$\begin{array}{rcl} X\times S & \longrightarrow & (X\times S)/(f\times g) \\ \mathrm{pr}_2\downarrow && \downarrow\pi\\ S& \longrightarrow & F \end{array}$$ is Cartesian, because both horizontal maps are étale double covers. Hence each section of $\pi$ lifts to a section of $\mathrm{pr}_2$, so determines (and is determined by) a morphism $S\to X$. Such a morphism must be constant (with image $x\in X$, say), and the compatibility with involutions implies that $x$ is a fixed point of $f$. So, $\pi$ has only finitely many sections, hence is not trivial.

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